Question: $\text B = \left[\begin{array}{rr}4 & 4 \\ 1 & 0 \\ -2 & 1\end{array}\right]$ and $\text F = \left[\begin{array}{rr}0 & 3 \\ 0 & 1\end{array}\right]$ Let $\text {H = BF}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{B}$ and the first column of $\text{F}$. $ \text {H}=\left[\begin{array}{rr}{4} & {4} \\ 1 & 0 \\ -2 & 1\end{array}\right]\left[\begin{array}{rr} {0} & 3 \\ {0} & 1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(4,4)\cdot(0,0)\\\\ &=4 \cdot 0 + 4\cdot 0\\\\ &=0 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4 \cdot 3 + 4\cdot 1 = 16$ (Choice B) B $1 \cdot 3 + 0\cdot 1 = 3$ (Choice C) C $1 \cdot 0 + 0\cdot 0 = 0$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}0 & 16 \\ 0 & 3 \\ 0 & -5\end{array}\right]$